3 Tutor System
Starting just at 265/hour

# In each of the following find the value of ‘k’, for which the points are collinear. (i) (7, -2), (5, 1), (3, -k) (ii) (8, 1), (k, -4), (2, -5)

We know that for collinear points area of triangle = 0 ,i.e.,
$$0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)]$$

(i) $$7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0$$
=> $$7 - 7k + 5k + 10 - 9 \ = \ 0$$
=> $$2k \ = \ 8$$
=> $$k \ = \ 4$$

(ii) $$8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0$$
=> $$8 - 6k + 10 \ = \ 0$$
=> $$6k \ = \ 18$$
=> $$k \ = \ 3$$